FALSE: every finite group is a direct product of cyclic prime-power groups

Given: The objects and hypotheses in the statement.

Let $X$ be a set. A permutation of $X$ is a bijection $f:X\to X$ (def-injection-surjection-bijection). The symmetric group of $X$ is the set of all permutations of $X$, $\mathrm{Sym}(X):=\{f:X\to X:f\text{ is a bijection}\},$ equipped with composition as its operation, $(f\circ g)(x)=f(g(x))(x\in X),$ and with the identity map ${\mathrm{i}\mathrm{d}}_X$, given by ${\mathrm{i}\mathrm{d}}_X(x)=x$, as distinguished element. Composition of two bijections of $X$ is again a bijection of $X$ (def-injection-surjection-bijection), so $\mathrm{Sym}(X)$ is closed under $\circ$ and $\circ$ is a binary operation on it (def-binary-operation); and ${\mathrm{i}\mathrm{d}}_X$ is a bijection of $X$, so it is an element of $\mathrm{Sym}(X)$, and it is a two-sided identity for composition (def-identity-element) because $f\circ {\mathrm{i}\mathrm{d}}_X=f={\mathrm{i}\mathrm{d}}_X\circ f$ holds pointwise for every $f$. That $(\mathrm{Sym}(X),\circ ,{\mathrm{i}\mathrm{d}}_X)$ is a group is lem-symmetric-group-is-a-group. Cycle notation for a finite list of distinct points. For distinct elements $x_0,x_1,\dots ,x_{k-1}$ of $X$ with $k\ge 2$, the symbol $(x_0{x}_1\cdots x_{k-1})$ denotes the permutation sending $x_i$ to $x_{i+1}$ for $i<k-1$, sending $x_{k-1}$ to $x_0$, and fixing every element of $X$ outside $\{x_0,\dots ,x_{k-1}\}$. It is a bijection, because the map described sends the set $\{x_0,\dots ,x_{k-1}\}$ onto itself by a rule with an evident inverse (send each $x_{i+1}$ back to $x_i$ and $x_0$ back to $x_{k-1}$) and fixes the complement pointwise. A transposition is such a symbol with $k=2$, that is $(ab)$ with $a\mathrm{\ne }b$: it exchanges $a$ and $b$ and fixes everything else, and it satisfies $(ab)\circ (ab)={\mathrm{i}\mathrm{d}}_X$. A product of cycle symbols means their composite, so $(ab)(cd)$ is $(ab)\circ (cd)$. (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition).

For every set $X$, the triple $(\mathrm{Sym}(X),\circ ,{\mathrm{i}\mathrm{d}}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$. If $X$ contains three distinct elements $a$, $b$, $c$, then $\mathrm{Sym}(X)$ is not abelian: the transpositions $\tau =(ab)$ and $\rho =(bc)$ satisfy $\tau \circ \rho \mathrm{\ne }\rho \circ \tau$. ($\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $A$ be a finite set with $n:=\mid A\mid$ and write $\mathrm{Bij}(A):=\{f:A\to A:f\text{ is a bijection}\}\mathrm{.}$ Then $\mathrm{Bij}(A)$ is finite and $\mid \mathrm{Bij}(A)\mid =n!$ (def-factorial-and-falling-factorial). More generally, for finite sets $X$ and $Y$ write $\mathrm{Bij}(X,Y)$ for the set of bijections $X\to Y$. If $\mid X\mid =\mid Y\mid =n$ then $\mathrm{Bij}(X,Y)$ is finite with $n!$ elements, and if $\mid X\mid \mathrm{\ne }\mid Y\mid$ then $\mathrm{Bij}(X,Y)=\mathrm{\varnothing }$. (A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Let $G$ be a group and $g\in G$, with integer powers as in def-group-power. Then $⟨g⟩=\{g^n:n\in \mathbb{Z}\},$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$. Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ($⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian).

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,h\in H\}$ and componentwise operation $(g,h)(g^{\mathrm{\prime }},h^{\mathrm{\prime }}):=(g{g}^{\mathrm{\prime }},h{h}^{\mathrm{\prime }})\mathrm{.}$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h{)}^{-1}=(g^{-1},h^{-1})\mathrm{.}$ Moreover the coordinate maps ${\pi }_G(g,h)=g$ and ${\pi }_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).